2010: Problem 23

Topic: Fluids
Concepts: General Newton’s 2nd law, Statics


Solution:

First, we study the force on one U-tube:

To find the force the stream of water exerts on the U-tube, we consider a small time interval Δt\Delta t where a small mass Δm\Delta m of water enters from the left at the top of the U-tube. Since the water is flowing at velocity vv, the length of this segment is ΔL=vΔt\Delta L=v\Delta t so

Δm=ρΔV=ρAΔL=ρAvΔt\Delta m=\rho\Delta V=\rho A\Delta L=\rho Av\Delta t

At the same time, we have Δm\Delta m of water leaving to the left with velocity v-v at the bottom of the U-tube. Thus, in time Δt\Delta t, water of mass Δm\Delta m effectively has its velocity changed from vv to v-v. This change in momentum can be attributed to the force from the U-tube on the water,

F=ΔpΔt=(Δm)v(Δm)(v)Δt=2(Δm)vΔt=2ρAv2F=\frac{\Delta p}{\Delta t}=\frac{(\Delta m)v-(\Delta m)(-v)}{\Delta t}=2\frac{(\Delta m)v}{\Delta t}=2\rho Av^2

By Newton’s 3rd law, this is also the force the water exerts on the U-tube. Going to the tube assembly of two U-tubes, since the net force is zero, we balance forces from each side:

F1=2ρAv2=2ρAv2=F2F_1=2\rho Av^2=2\rho A’v’^2=F_2
v=vAA=2vv’=v\sqrt{\frac{A}{A’}}=\sqrt{2}v

so the answer is C.

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